Osmotic pressure –

source : study.com

### Solution )

0.1 M  means 0.1 mole of solute in 1 litre of solution.

π   =  ? atm  , S = 0.0821 litre atm K -1mole-1 , T  = 18 0 C = 273 + 18 =291 K , n = 0.1 mole  , V = 1 litre

π N = n ST / V

π N = 0.1  x  0.0821 x 291 / 1

π N=  2.389 atm.

K4 [Fe (CN)6]                 ——–>   4 K +        +     [Fe (CN)6] 4-

mole before dissociation                     1                                                  0                       0

moles after  dissociation                    1 – ∝                                          4 ∝                       ∝

Total moles after  dissociation  = 1 – ∝ + 4∝ + ∝ = 1 + 4∝

% of ∝ = 46 % , ∝ = 46 / 100 = 0.46

π  exp / π N = (1 + 4 x 0.46 ) / 1

π exp /  2.389 = ( 1 + 4 x 0.46 )

π exp =   2.389 x 2.84

π exp =  6.78 atm  Ans.

### Solution )

C (molarity ) = ?

T =300 K , S = 0.0821 Litre atm K-1 mole-1,

π  = 2.0 atm

π  = CST

C = 2.0 / 0.0821 x 300 = 0.0812 mole / litre

In dilute solution , the density of water = 1 gm / cc

ΔTf  = Kf x molality

ΔTf = 1.86 x 0.0812

ΔTf  = 0.151

freezing point of solvent (water ) = 0 + 273 = 273 K

ΔTf = freezing point of solvent –   freezing point of solution

freezing point of solution = 273 – 0.151

freezing point of solution = 272.749 K

OR

freezing point of solvent (water ) = 00 C

ΔTf = freezing point of solvent –   freezing point of solution

freezing point of solution = 0 – 0.151

freezing point of solution =  – 0.151 0C  Ans.

### Solution )

For ‘X’ ,

w1 = 0.12 gm , v1 = 1 litre

π1 = w1 S T / m1 v1

π 1 = 0.12 S T / m1 x 1

For ‘Y’ ,

w2 =0.18 gm , v2 = 1 litre

π2 = w2 S T / m2 v2

π2 = 0.18 ST /m2 x 1

π 2 = 0.12 m/ 0.18 m1

2 / 1 = 0.12 m/ 0.18 m1

m2 / m1= 2 x 0.18 / 0.12

m2 / m1 = 3 /1

m1 : m2 = 1 : 3  Ans.

### Solution )

π   =  ? atm  , S = 0.0821 litre atm K -1mole-1 , T  = 270 C = 273 + 27 = 300 K ,

Molecular weight of NaCl =23 + 35.5 = 58.5

Equivalent weight of NaCl = Molecular weight / total positive or negative valency

total positive valency  = 1 (because Na+)

Equivalent weight of NaCl = 58.5  / 1 = 58.5

Molarity / normality =   Equivalent weight/ molecular weight

M / N = 58.5 / 58.5

N = M

Decimolar solution means ,   M = 0.1,

Therefore,                            N =0.1

C is molarity of solution , C =0.1,

π N = C ST

π N = 0.1  x  0.0821 x 300= 2.46 atm

π N=  2.46 atm.

NaCl              ——–>                Na+        +     Cl

mole before dissociation                     1                                                  0                       0

moles after  dissociation                    1 – ∝                                           ∝                       ∝

Total moles after  dissociation  = 1 – ∝ + ∝ + ∝ = 1 + ∝

% of ∝ = 80 % , ∝ = 80 / 100 = 0.80

π  exp / π N = (1 +  0.80) / 1

π exp /  2.46 = 1.80

π exp =   2.46 x 1.80

π exp =  4.43 atm  Ans.